A121437 Matrix inverse of triangle A122177, where A122177(n,k) = C( k*(k+1)/2 + n-k + 2, n-k) for n>=k>=0.

1, -3, 1, 6, -4, 1, -16, 14, -6, 1, 63, -62, 33, -9, 1, -351, 365, -215, 72, -13, 1, 2609, -2790, 1731, -642, 143, -18, 1, -24636, 26749, -17076, 6696, -1664, 261, -24, 1, 284631, -311769, 202356, -81963, 21684, -3831, 444, -31, 1, -3909926, 4305579, -2822991, 1166310, -320515, 60768, -8012, 713, -39, 1

Offset

0,2

Links

Table of n, a(n) for n=0..54.

Formula

(1) T(n,k) = A121436(n-1,k) - A121436(n-1,k+1). (2) T(n,k) = (-1)^(n-k)*[A107876^(k*(k+1)/2 + 3)](n,k); i.e., column k equals signed column k of A107876^(k*(k+1)/2 + 3). G.f.s for column k: (3) 1 = Sum_{j>=0} T(j+k,k)*x^j/(1-x)^( j*(j+1)/2) + j*k + k*(k+1)/2 + 3); (4) 1 = Sum_{j>=0} T(j+k,k)*x^j*(1+x)^( j*(j-1)/2) + j*k + k*(k+1)/2 + 3).

From Benedict W. J. Irwin, Nov 26 2016: (Start)

Conjecture: The sequence (column 2 of triangle) 14, -62, 365, -2790, 26749, ... is described by a series of nested sums:

14 = Sum_{i=1..4} (i+1),

-62 = -Sum_{i=1..4} (Sum_{j=1..i+1} (j+2)),

365 = Sum_{i=1..4} (Sum_{j=1..i+1} (Sum_{k=1..j+2} (k+3))),

-2790 = -Sum_{i=1..4} (Sum_{j=1..i+1} (Sum_{k=1..j+2} (Sum_{l=1..k+3} (l+4)))). (End)

Example

Triangle begins:
1;
-3, 1;
6, -4, 1;
-16, 14, -6, 1;
63, -62, 33, -9, 1;
-351, 365, -215, 72, -13, 1;
2609, -2790, 1731, -642, 143, -18, 1;
-24636, 26749, -17076, 6696, -1664, 261, -24, 1;
284631, -311769, 202356, -81963, 21684, -3831, 444, -31, 1; ...

Prog

(PARI) /* Matrix Inverse of A122177 */ T(n, k)=local(M=matrix(n+1, n+1, r, c, if(r>=c, binomial((c-1)*(c-2)/2+r+1, r-c)))); return((M^-1)[n+1, k+1])
(PARI) /* Obtain by g.f. */ T(n, k)=polcoeff(1-sum(j=0, n-k-1, T(j+k, k)*x^j/(1-x+x*O(x^n))^(j*(j+1)/2+j*k+k*(k+1)/2+3)), n-k)

Crossrefs

Cf. A098568, A107876, A122177, A107885.

Sequence in context: A210230 A207615 A209165 * A078585 A116551 A213783

Adjacent sequences: A121434 A121435 A121436 * A121438 A121439 A121440

Keyword

sign,tabl

Author

Paul D. Hanna, Aug 27 2006

Status

approved